I have been trying to prove Th1.9 rigorously since Friedberg didn't do so. Here is my attempt at a rigorous proof.
Th 1.9 Restated: If a vector space V, S has n elements and span(S)=V then some subset of S is a basis for V.
Proof:
We proceed by induction on n.
Base step n=1: Obvious.
Inductive Step: Inductive Hypothesis: Suppose $\forall V \forall H $ If a vector space V, H has n elements and span(H)=V then some subset of H is a basis for V.
Suppose that V is a vector space, S has n+1 elements and span(S)=V. If S is linearly independent then we are done. If S is linearly dependent (here is where I get stuck).
I want to say to remove elements from S until it is linearly independent than this new set, lets say S', will be subset of S with span(S')=Span(S)=V thus we proved the theorem. But doing it this way is not rigorous which defeats the whole purpose. Is there any help you can offer ?
Also am I right in saying that Friedberg's proof is non-rigorous because of the sentence "Continue if possible choosing vectors $u_2,...u_k$ in S such that $\{u_1,...,u_k\}$ is linearly independent. Since S is a finite set, we must eventually reach a stage at which $B=\{u_1,...,u_k\}$ is linearly independent subset of S, but adjoining to B any vector in S not in B produces a linearly dependent set. I feel this statement is way too informal and not rigorous. Am I right ?
The proof is rigorous. By theorem 1.7, if there exists a vector $u_2 \in S$ such that $u_2 \notin \textrm{span}(\{u_1\})$, then $\{u_1,u_2\}$ is a linearly independent set. We can keep adding vectors in this way until we reach a maximal set $\{u_1,\ldots,u_k\}$ such that if we have another other vector $v\in S$, $v\in\textrm{span}(\{u_1,\ldots,u_k\})$. We will eventually reach this maximal set, because even if we run out of vectors in $S$, that would imply $S$ is linearly independent, so it would be a basis for $V$.